Question: $f(x, y) = \dfrac{x^2}{\sqrt{x^2 + y^2}} - x^2 - y^2$ We have a change of variables: $\begin{aligned} x &= X_1(r, \theta) = r \cos(\theta) \\ \\ y &= X_2(r, \theta) = r \sin(\theta) \end{aligned}$ What is $f(x, y)$ under the change of variables? Choose 1 answer: Choose 1 answer: (Choice A) A $\cos^2(\theta) - r^2$ (Choice B) B $r^2\cos^2(\theta)$ (Choice C) C $-r^2\sin^2(\theta)$ (Choice D) D $r\cos^2(\theta) - r^2$
When applying a change of variables, we substitute the new definition for $x$ and $y$ into the original equation. The original equation: $f(x, y) = \dfrac{x^2}{\sqrt{x^2 + y^2}} - x^2 - y^2$ Let's substitute $X_1(r, \theta)$ for $x$ and $X_2(r, \theta)$ for $y$. We can rewrite every $x^2 + y^2$ using the trigonometric identity that $\cos^2(\theta) + \sin^2(\theta) = 1$ : $\begin{aligned} x^2 + y^2 &= r^2\cos^2(\theta) + r^2\sin^2(\theta) \\ \\ &= r^2(\cos^2(\theta) + \sin^2(\theta)) \\ \\ &= r^2 \end{aligned}$ Therefore: $\begin{aligned} f(x, y) &= \dfrac{r^2\cos^2(\theta)}{\sqrt{r^2}} - r^2 \\ \\ &= \dfrac{r^2\cos^2(\theta)}{r} - r^2 \\ \\ &= r\cos^2(\theta) - r^2 \end{aligned}$ Therefore, under the change of variables, $f(x, y)$ becomes: $r\cos^2(\theta) - r^2$